Simplify; express your answer in exponential form. Assume $t\neq 0, r\neq 0$. $\dfrac{{(t^{-1})^{-3}}}{{(t^{2}r^{-5})^{5}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${t^{-1}}$ to the exponent ${-3}$ . Now ${-1 \times -3 = 3}$ , so ${(t^{-1})^{-3} = t^{3}}$ In the denominator, we can use the distributive property of exponents. ${(t^{2}r^{-5})^{5} = (t^{2})^{5}(r^{-5})^{5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(t^{-1})^{-3}}}{{(t^{2}r^{-5})^{5}}} = \dfrac{{t^{3}}}{{t^{10}r^{-25}}}$ Break up the equation by variable and simplify. $\dfrac{{t^{3}}}{{t^{10}r^{-25}}} = \dfrac{{t^{3}}}{{t^{10}}} \cdot \dfrac{{1}}{{r^{-25}}} = t^{{3} - {10}} \cdot r^{- {(-25)}} = t^{-7}r^{25}$.